Question

Of 10 girls in a class, three have blue eyes. Two of the girls are chosen at random. Find the probability that: (a) both have blue eyes; (c) at least one has blue eyes; (b) neither has blue eyes; (d) exactly one has blue eyes.

Answer 1

The probabilities of selecting girls with blue eyes from a class are calculated for different scenarios: both having blue eyes, neither having blue eyes, at least one having blue eyes, and exactly one having blue eyes, using fundamental probability rules.

Step-by-step explanation:

The question pertains to the rules of probability within the context of selecting girls with a specific eye color from a class. Let's address each part of the question:

1. (a) both have blue eyes: The probability of the first girl having blue eyes is 3/10. Since one girl with blue eyes is now taken, the probability that the second girl also has blue eyes is 2/9. Use the multiplication rule of probability: (3/10) × (2/9) = 1/15.
2. (b) neither has blue eyes: The probability that the first girl does not have blue eyes is 7/10 (since 7 girls do not have blue eyes), and the probability that the second girl chosen also does not have blue eyes (now out of 9 remaining girls) is 6/9 or 2/3. Thus, (7/10) × (2/3) = 7/15.
3. (c) at least one has blue eyes: This is the complement of (b), so it's 1 - P(neither has blue eyes). Calculating it gives us 1 - 7/15 = 8/15.
4. (d) exactly one has blue eyes: This can occur in two ways: blue-eyed girl first then not, or not then blue-eyed girl. (3/10) × (7/9) + (7/10) × (3/9) = 21/90 + 21/90 = 7/30.

We calculated the probabilities of all possible outcomes regarding the selection of girls with blue eyes based on combinatorial principles and fundamental probability laws.

Answer 2

C.

Step-by-step explanation:

It's the most reasonable answer.